Question: Solve for $x$ : $3x^2 - 36x + 60 = 0$
Answer: Dividing both sides by $3$ gives: $ x^2 {-12}x + {20} = 0 $ The coefficient on the $x$ term is $-12$ and the constant term is $20$ , so we need to find two numbers that add up to $-12$ and multiply to $20$ The two numbers $-2$ and $-10$ satisfy both conditions: $ {-2} + {-10} = {-12} $ $ {-2} \times {-10} = {20} $ $(x {-2}) (x {-10}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -2) (x -10) = 0$ $x - 2 = 0$ or $x - 10 = 0$ Thus, $x = 2$ and $x = 10$ are the solutions.